Problem: Multiply the following complex numbers: $({i}) \cdot ({-4-4i})$
Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({i}) \cdot ({-4-4i}) = $ $ ({0} \cdot {-4}) + ({0} \cdot {-4}i) + ({1}i \cdot {-4}) + ({1}i \cdot {-4}i) $ Then simplify the terms: $ (0) + (0i) + (-4i) + (-4 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 0 + (0 - 4)i - 4i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 0 + (0 - 4)i - (-4) $ The result is simplified: $ (0 + 4) + (-4i) = 4-4i $